Question

Prove that the sufficent condition for the alignment of three points in a bidimensional space is:

Answer 1

\[\left[\begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1\\ x_3 & y_3 & 1\end{matrix}\right] = 0\]

Answer 2

Sufficient condition for alignment of 3 pts in 2D is that the slope of segment (x1,y1)-(x2,y2) is the same as that of segment (x2,y2)-(x3,y3)

So (y1-y2)/(x1-x2)=(y3-y2)/(x3-x2)
rearrange above: (y1-y2)(x3-x2) - (y3-y2)(x1-x2) = 0
expand, simplify n re-group, u will get the determinant form above.

Answer 3

Well, I thought of, having point x1 as the tail of the vectors: x1x2 and x1x3 and then that would mean that the vectors would have coordinates ((x2-x1),(y2-y1)) and (x3-x1, y3-y1)
So, since they must be parallel:
\[\left[\begin{matrix}(x_2-x_1)& (y_2 - y_1) \\ (x_3 - x_1) & (y_3 - y_1)\end{matrix}\right]=0\]
And here, I thought I cound just rewrite that determinant as:
\[\left[\begin{matrix}(x_1) & (y_1) & 1 \\ (x_2-x_1) & (y_2-y_1) & 1\\ (x_3-x_1) & (y_3-y_1) & 1\end{matrix}\right]=0\]

I'm not sure what transformation...

Answer 4

Right idea but it would be difficult to go from a 2x2 to a 3x3 determinant. So what I did was to expand the eqn for slopes. Then expand the determinant form to its individual terms. They will match up with the slope eqn. They did for me :)

Answer 5

Ah... I see, I did it that way and I got the result.
I was too keen to work with determinants that I did not look for that method. Thanks!

Answer 6

hahahaa i HATE determinants so i rather expand one out if i can. thanx 4 the medal, @Owlcoffee

Answer 7

interesting...